408算法练习——左叶子之和

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode() {}
 8  *     TreeNode(int val) { this.val = val; }
 9  *     TreeNode(int val, TreeNode left, TreeNode right) {
10  *         this.val = val;
11  *         this.left = left;
12  *         this.right = right;
13  *     }
14  * }
15  */
16 class Solution {
17     public int sumOfLeftLeaves(TreeNode root) {
18         if(isleaf(root)){
19             return 0;
20         }
21         return dfs(root);
22     }
23     public int dfs(TreeNode root){
24         int ans = 0;
25         if(root.left != null){
26             ans += isleaf(root.left)?root.left.val:dfs(root.left);
27         }
28         if(root.right !=null && !isleaf(root.right)){
29             ans += dfs(root.right);
30         }
31         return ans;
32     }
33     public boolean isleaf(TreeNode node){
34         return node.left == null && node.right == null;
35     }
36 }